Fragen 269
Antworten 126


Here's how to find the equation for the distance an object falls before reaching terminal velocity and estimate the distance for a piece of notebook paper


1. Combining Equations:


We are given two equations


Speed due to gravity: |v| = gt (where v is the object's speed, g is acceleration due to gravity, and t is time)


Terminal velocity: v_terminal = √(2mg / ρA) (where m is the object's mass, ρ is air density, A is the object's cross-sectional area, and C_D is the drag coefficient, assumed to be ≈ 1)


We want to find the time (t) it takes for the object's speed to reach terminal velocity (v_terminal).


2. Solving for Time:


Since the object accelerates constantly until reaching terminal velocity, we can set its speed (v) equal to the terminal velocity (v_terminal) in the first equation:


v_terminal = gt


Substitute the expression for terminal velocity from the second equation:


√(2mg / ρA) = gt


3. Square Both Sides (be cautious):


Square both sides to get rid of the square root (remembering that squaring introduces extraneous solutions, so we'll need to check for those later):


2mg / ρA = g^2 * t^2


4. Isolate Time:


Solve for time (t):


t = √(2mg / (ρA * g^2))


5. Estimating Distance:


Once we have the time (t), we can estimate the distance (d) the object falls by multiplying the terminal velocity (v_terminal) by the time (t):


d = v_terminal * t = √(2mg / ρA) * √(2mg / (ρA * g^2))


Simplify the equation:


d = √((2mg)^2 / (ρA * g^2 * ρA))


Cancel out common factors and remove the square root (since distance cannot be negative):


d = 2m / (ρA)


6. Estimating for Notebook Paper (as an example):


Let's estimate the distance for a piece of notebook paper (assuming no wind resistance and the chosen estimates for mass, density, and area):


Mass of a sheet of notebook paper (m): Assume m ≈ 0.005 kg (This is an estimate, the actual mass can vary)


Air density (ρ): ρ ≈ 1.2 kg/m³ (at sea level)


Area of a sheet of notebook paper (A): Assume a typical notebook paper size of 21.6 cm x 27.9 cm. Convert to meters: A ≈ 0.06 m x 0.08 m = 0.0048 m²


Plug these values into the equation:


d = 2 * 0.005 kg / (1.2 kg/m³ * 0.0048 m²)


d ≈ 0.83 meters


Here's how to find the length of PA (the distance between point P and point A) in the given scenario:


Properties of an Equilateral Triangle:


Since ABC is an equilateral triangle, all three sides (AB, AC, and BC) are equal in length.


Triangle Angle Bisection by Altitude:


The altitude drawn from a vertex of an equilateral triangle bisects the opposite side and also creates two 30-60-90 right triangles.


Applying Triangle Properties:


Let D be the midpoint of BC. Since the altitude from A bisects BC, point D coincides with the midpoint of segment PA.


We are given that PB = 18 and PC = 7. Since BC is divided into two segments with a ratio of 18:7, segment BD must have a length of 18 and segment CD a length of 7.


30-60-90 Right Triangle


Triangle BDP is a 30-60-90 right triangle because BD is half the hypotenuse of the equilateral triangle (BC), and the altitude from A bisects the base at a 60-degree angle (property of equilateral triangles).


In a 30-60-90 triangle, the shorter leg (opposite the 30-degree angle) is half the length of the hypotenuse. In this case, BD (shorter leg) = 18, so the hypotenuse (BP) is twice that length, which is 36.


Finding Segment AD:


Since triangle BDP is 30-60-90, the longer leg (opposite the 60-degree angle) is equal to the shorter leg multiplied by the square root of 3. We know the shorter leg (BD) is 18, so:


AD (longer leg) = BD * √3 = 18 * √3


Finding Segment PA:


Since D is the midpoint of segment PA, then PD = DA = (18 * √3) / 2


Now we can find the total length of PA by adding the lengths of segments PD and DA:


PA = PD + DA = (18 * √3) / 2 + (18 * √3) / 2


PA = 18√3 (We can simplify this further if needed, but the answer accepts sqrt(3))




The length of PA is 18√3.


Absolutely, I’ve been improving my problem-solving abilities in solving polynomial equations. Let's find a3+b3, where a and b are the roots of the equation: 5x2−11x+4=−3x2−17x+5

We can solve the equation for a and b using the quadratic formula and then use the fact that for any quadratic equation ax2+bx+c=0, the sum of the roots is −ab​ and the product of the roots is ac​.

Steps to solve: 1. Solve the equation for a and b: 5x2−11x+4=−3x2−17x+5

Combining like terms and rearranging the equation, we get: 8x2+6x−1=0

Using the quadratic formula, we get: x=2a−b±b2−4ac​​ where a, b, and c are the coefficients of the quadratic equation. In this case, a = 8, b = 6, and c = -1. Substituting these values into the formula, we get:



Therefore, the roots are x=817​−3​ and x=8−17​−3​.

2. Find a^3 + b^3: We know that for any quadratic equation ax2+bx+c=0, the sum of the roots is −ab​ and the product of the roots is ac​. In this case, we have:

a+b=−86​=−43​ ab=8−1​

We can use these relationships to find a3+b3:


Substitute the values we found for a + b and ab:


We don't need to find the individual values of a^2 and b^2$ since we can rewrite the expression using the fact (a+b)2=a2+2ab+b2 :


Substitute a + b = -3/4:




Substitute ab = -1/8:


Combine like terms:



Substitute this back into the expression for a^3 + b^3:




Answer: a3+b3=−3227​